Originally Posted by

**katie** Hi guys,

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2

Mr F says: $\displaystyle {\color{red} \frac{u^2 + 1}{u^2 - 4u + 4} = \frac{(u^2 - 4u + 4) + (4u - 3)}{u^2 - 4u + 4} = 1 + \frac{4u - 3}{u^2 - 4u + 4}}$.

Now note that $\displaystyle {\color{red} \frac{4u - 3}{u^2 - 4u + 4} = \frac{(4u - 8) + 5}{u^2 - 4u + 4} = \frac{4u - 8}{u^2 - 4u + 4} + \frac{5}{u^2 - 4u + 4}}$.

Therefore $\displaystyle {\color{red} \frac{u^2 + 1}{(u-2)^2} = 1 + \frac{5}{(u-2)^2} + \frac{4u - 8}{u^2 - 4u + 4}}$.

See what you can do with that.

and the antiderivative of cosxe^(sinx) dx

Mr F says: Make the substitution $\displaystyle {\color{red} u = \sin x}$.

It asks to find the volume of the solid of revolution obtained by rotating, a full turn about the x-axis, the area between the x-axis and the curve y = 2 sin x, for x ∈ [ π , 3π ].

Mr F says: By symmetry, it'll be twice the volume when the area from 0 to pi is rotated:

$\displaystyle {\color{red} V = 2 \pi \int_{0}^{\pi} (2 \sin x)^2 \, dx = ......}$.

Thnanks